Reaction at the Interfaces

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1 Reaction at the Interfaces Lecture 1

2 On the course Physics and Chemistry of Interfaces by HansJürgen Butt, Karlheinz Graf, and Michael Kappl Wiley VCH; 2nd edition (2006)

3 Interfaces interface the region where properties change from one phase to another dispersed d phase (colloid) important t case for interface science as the properties are large determined by interfaces dispersed phase continuous phase in some cases dispersed and continuous phases can be difficult to distinguish (intervowen phases)

4 Types of interfaces it s possible to classify interfaces based on the nature of bulk phase. Gases intermix completely, so there are no gas-gas interface gas-liquid G-L fluid interfaces liquid1-liquid2 q L-L 1 2 gas-solid G-S solid interfaces liquid-solid L-S solid-solid S 1 -S 2

5 Key concepts Surface tension Wetting Adsorption Emulsions Colloids Membranes

water) Different concentration of solutes (adsorption) The thickness of

6 Why the surface is different Different density Different orientation of molecules (e.g. water) Different concentration of solutes (adsorption) The thickness of the interface will be different when defined from different viewpoints (e.g. thinner in terms density and larger in terms of adsorption)

7 Stability of an interface interface can possess and extra energy, so G = γ A+other terms surface tension should be positive surface tension should be positive otherwise the system is totally miscible

Surface tension Surface tension can be defined as a force per unit length acting on an interface F γ = δ x Forces can be understood as a result of broken

8 Surface tension Surface tension can be defined as a force per unit length acting on an interface F γ = δ x Forces can be understood as a result of broken bonds when moved to a dissimilar phase. Rough estimate of a bond energy (Example 2.3) gives a reasonable values work of extension: w = Fδ y = γxδy = γδa s

9 Effect of temperature on surface tension experimentally, surface tension of pure liquids drops linearly with the temperature Eötvös equation: d 2/3 ( γ ( M / ρ) ) dt = Jmol K 7 2/3 1

10 Young-Laplace equation If the surface is curved in equilibrium, there should be a pressure difference across it Example: Rubber membrane on a tube The Young-Laplace equation: 1 1 Δ P = γ + R R If the shape of a surface is known the pressure difference can be found in the absence of external fields the pressure is the same everywhere so the curvature should be the constant as well If the pressure difference is known, the curvature can be calculated 1 2 should be the constant as well

11 The Laplace equation pressure forces: α β δf = ( P P ) δacosφ α β F = ( P P ) π rc surface tension: 2 c γ F = γ(2 πr )cos θ = γ(2 πr ) r / r z c c c α 2 ( P P β ) πrc γ(2 πrc) rc / r = 0 2 γ α β P P = Laplace equation for spherical surface r

12 in general case: The Laplace equation π r /2 α β 2 2r 2r P P πr = γdl + R 0 1 R2 ( ) P P = γ + = r r rm α β γ r P 1 = P 2 for large structure, gravity should be taken into account: 1 1 Δ P = γ + + ρ gh R1 R2

13 Measurements of surface tension From the shape (contour) measurements sessile drops pendent drops pendent bubbles sessile bubbles

14 Measurements of surface tension Drop weight method mg = 2π r C γ mg = 2π frc γ Neck formation in practice a correction factor is introduced to take inro p account neck formation

15 Measurements of surface tension Maximum bubble pressure γ = rc ΔP /2 Valving effect in capillaries 4γ cosθ 4γ 4γ cosθ Δ P = Δ P = Δ P = d1 d1 d2

16 Measurements of surface tension Drop weight (pendent droplet) mg = 2π rγ Neck formation Correction factor required due to neck formation

17 Measurements of surface tension Wilhelmy plate: F = γ 2( x+ y) in the past was mainly measured on roughened mica, etched glass etc. Currently paper plates (i.e. filter paper) is the material of choice du Noüy ring: F = 2 πγ ( r + r ) i a

18 Measurements of surface tension Capillary rise γ π ρ π 2 2 r c =Δ gh r c 1 γ = Δρghr c 2 alternatively the difference between two capillaries of different diameter can be measured Dynamic methods: relaxation of an elliptic liquid jet.

19 The Kelvin equation vapour pressure above a droplet c p ln = p L γv 2 RT r m Kelvin equation dg = SdT + Vdp = G = Vm, μ m μ p α β α α β β δμ = δμ V δ p = V δ p β β α α δ p ( V V )/ V = 2 γδ (1/ r m ) α β δ p δ p = 2 γδ(1/ rm ) β α if V V, δ pv β β / V α = δμ β / V α = 2 γδ(1/ r m ) c L μ μ = 2 γv (1/ r m ) c L p γv 2 RT ln = p RT rm T

20 Consequences of Kelvin equation smaller droplets will have higher vapour pressure and therefore evaporate faster small droplets have higher chemical potential small bubbles have lower vapour pressure condensation in a capillary at the phase transition only the nucleation center with infinite radius will grow. All finite nucleation center require finite overcooling/overheating (i.e. a thermodynamic force)

21 Capillary condensation condensation in a capillary (e.g. in a porous materials) will happen below the dew point RT ln k P 0 = P 0 2γV r C m in the case of contact angle ϴ r = r C cos Θ

22 Capillary adhesion of fine particles capillary condensation will cause adhesion between particles in a powder: F = πx γ πx γ x r r ( ) ( ) F p + = + + p ;2 p = + 2 R r x r R rr x xr x the force is independent on the vapour pressure and curvature radius: their effect is mutually compensated! 2 π RP γ

23 Nucleation theory Homogeneous nucleation: nucleation in the absence of an external surface. As a phase transformation requires formation of an interface, the change of Gibbs free energy is: 3 4π r 2 Δ G = Δ μ+ 4 π r γ 3V m 2 4π r Δ G = Δ μ + 8 π r γ * r = 2γVm r Vm Δμ In the case of vapour condensation: μ = μ + RT ln P μ = μ + RT ln P k V Δ μ = 0 L 0 0 k ( 0 ) RT ln P P 3 4π r k 2 Δ G = RTln ( P0 P) + 4π r γ 3V m * 2γVm r = k RT ln P P ( 0 )

24 Probelms Problem 1 A jet aircraft is flying through a region where the air is 10% supersaturated t with water vapour (i.e. the relative humidity is 110%). After cooling, the solid smoke particles emitted by the jet engines adsorb water vapour and can then be considered as minute spherical droplets. What is the minimum radius of these droplets if condensation is to occur on them and a vapour trail form? Data: γ(h 2 O) = 75.2 mn m -1, M(H 2 O) = kg mol -1, ρ(h 2 O) = 1030 kg m -3, T = 275 K. Problem 2. A hydrophilic sphere of radius R p = 5 µm sits on a hd hydrophilic hili planar surface. Wt Water from the surrounding atmosphere condenses into the gap. What is the circumference cu e ce of the meniscus? Make a plot of radius of circumference x versus humidity. At equilibrium the humidity is equal to P 0K /P 0

Surface chemistry. Liquid-gas, solid-gas and solid-liquid surfaces.

Surface chemistry. Liquid-gas, solid-gas and solid-liquid surfaces. Levente Novák & István Bányai, University of Debrecen Dept of Colloid and Environmental Chemistry http://kolloid.unideb.hu/~kolloid/